Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int a;
struct Result {
  int n;
  int value[10];
  void add(const vector<int> &v) {
    vector<int> w(v);
    for (int i = 0; i < n; ++i) w.push_back(value[i]);
    sort(w.begin(), w.end());
    n = min(::a, (int)w.size());
    for (int i = 0; i < n; ++i) value[i] = w[i];
  }
  friend inline bool operator==(const Result &A, const Result &B) {
    if (A.n != B.n) return false;
    for (int i = 0; i < A.n; ++i)
      if (A.value[i] != B.value[i]) return false;
    return true;
  }
};
void COMBINE(const Result &A, const Result &B, Result *result) {
  result->n = min(::a, A.n + B.n);
  for (int a = 0, b = 0; a + b < ::a;) {
    if (a < A.n && (b >= B.n || A.value[a] < B.value[b])) {
      result->value[a + b] = A.value[a];
      ++a;
    } else {
      result->value[a + b] = B.value[b];
      ++b;
    }
  }
}
struct Node {
  vector<int> neigh;
  vector<int> values;
  int parent;
  int cid;
  int cpos;
  int size;
  int height;
  int anc[17];
} node[100004];
struct TourNode {
  TourNode *left, *right;
  Result result;
  int a, b;
} buffer[270000], *alokator = buffer;
struct Chain {
  vector<int> nodes;
  TourNode *tour_root;
} chain[100004];
int N, M, K, Q;
void REK_SIZE(int x, int parent) {
  node[x].height = parent == -1 ? 0 : node[parent].height + 1;
  node[x].anc[0] = parent;
  for (int i = 1; i < 17; ++i) {
    node[x].anc[i] =
        node[x].anc[i - 1] == -1 ? -1 : node[node[x].anc[i - 1]].anc[i - 1];
  }
  for (int y : node[x].neigh) {
    if (y == parent) continue;
    REK_SIZE(y, x);
    node[x].size += node[y].size;
  }
  ++node[x].size;
}
void REK(int x, int parent, int cid) {
  int best = -1;
  int best_size = 0;
  if (cid == -1) cid = K++;
  node[x].parent = parent;
  node[x].cid = cid;
  node[x].cpos = (int)chain[cid].nodes.size();
  chain[cid].nodes.push_back(x);
  for (int y : node[x].neigh) {
    if (y == parent) continue;
    if (node[y].size > best_size) {
      best_size = node[y].size;
      best = y;
    }
  }
  if (best == -1) return;
  REK(best, x, cid);
  for (int y : node[x].neigh) {
    if (y == parent || y == best) continue;
    REK(y, x, -1);
  }
}
TourNode *TOUR_INIT(int a, int b, vector<int> &nodes) {
  TourNode *p = alokator++;
  p->a = a;
  p->b = b;
  if (a != b) {
    p->left = TOUR_INIT(a, (a + b) / 2, nodes);
    p->right = TOUR_INIT((a + b) / 2 + 1, b, nodes);
    COMBINE(p->left->result, p->right->result, &p->result);
  } else {
    auto &values = node[nodes[a]].values;
    sort(values.begin(), values.end());
    p->result.n = min((int)values.size(), 10);
    for (int i = 0; i < p->result.n; ++i) p->result.value[i] = values[i];
  }
  return p;
}
int qa, qb;
Result qresult;
void TOUR_QUERY(TourNode *p) {
  if (p->a > qb || p->b < qa) return;
  if (qa <= p->a && p->b <= qb) {
    Result tmp = qresult;
    COMBINE(tmp, p->result, &qresult);
  } else {
    TOUR_QUERY(p->left);
    TOUR_QUERY(p->right);
  }
}
void INIT_CHAIN(int k) {
  chain[k].tour_root =
      TOUR_INIT(0, (int)chain[k].nodes.size() - 1, chain[k].nodes);
}
void QUERY_CHAIN(int cid, int a, int b) {
  qa = a;
  qb = b;
  return TOUR_QUERY(chain[cid].tour_root);
}
void QUERY_CHAINS(int x, int node_cnt) {
  while (node_cnt > 0) {
    int cid = node[x].cid;
    int cpos = node[x].cpos;
    int cposb = max(0, cpos - node_cnt + 1);
    QUERY_CHAIN(cid, cposb, cpos);
    node_cnt -= cpos - cposb + 1;
    if (node_cnt <= 0) break;
    x = node[chain[cid].nodes[0]].parent;
  }
}
int CLIMB(int a, int x) {
  for (int i = 0; (1 << i) <= x; ++i)
    if (x & (1 << i)) a = node[a].anc[i];
  return a;
}
int LCA(int a, int b) {
  if (node[a].height > node[b].height)
    a = CLIMB(a, node[a].height - node[b].height);
  else if (node[b].height > node[a].height)
    b = CLIMB(b, node[b].height - node[a].height);
  if (a == b) return a;
  for (int i = 16; i >= 0; --i) {
    int aa = node[a].anc[i];
    int bb = node[b].anc[i];
    if (aa == bb) continue;
    a = node[a].anc[i];
    b = node[b].anc[i];
  }
  return node[a].parent;
}
int BRUTE_LCA(int a, int b) {
  while (a != b) {
    if (node[a].height > node[b].height)
      a = node[a].parent;
    else if (node[b].height > node[a].height) {
      b = node[b].parent;
    } else {
      a = node[a].parent;
      b = node[b].parent;
    }
  }
  return a;
}
void QUERY(int a, int b) {
  int lca = LCA(a, b);
  if (node[a].height > node[b].height) swap(a, b);
  QUERY_CHAINS(a, node[a].height - node[lca].height);
  QUERY_CHAINS(b, node[b].height - node[lca].height + 1);
}
bool BRUTE_QUERY(int x, int parent, int target, Result *result) {
  if (x == target) {
    result->add(node[x].values);
    return true;
  }
  for (int y : node[x].neigh) {
    if (y == parent) continue;
    if (BRUTE_QUERY(y, x, target, result)) {
      result->add(node[x].values);
      return true;
    }
  }
  return false;
}
int main(void) {
  scanf("%d%d%d", &N, &M, &Q);
  for (int i = 1; i < N; ++i) {
    int a, b;
    scanf("%d%d", &a, &b);
    node[a].neigh.push_back(b);
    node[b].neigh.push_back(a);
  }
  for (int i = 0; i < M; ++i) {
    int x;
    scanf("%d", &x);
    node[x].values.push_back(i + 1);
  }
  ::a = 10;
  REK_SIZE(1, -1);
  REK(1, -1, -1);
  for (int i = 0; i < K; ++i) INIT_CHAIN(i);
  for (int i = 0; i < Q; ++i) {
    int v, u;
    scanf("%d%d%d", &v, &u, &::a);
    qresult.n = 0;
    QUERY(v, u);
    printf("%d", qresult.n);
    for (int i = 0; i < qresult.n; ++i) printf(" %d", qresult.value[i]);
    printf("\n");
  }
  return 0;
}